# Volume form in terms of the Levi-Civita Symbol

I'm currently reading Sean Carroll's book on General Relativity, and at some point he writes:

First notice that the definition of the wedge product allows us to write $$\begin{equation} \mathrm{d}x^0\wedge\cdots\wedge\mathrm{d}x^{n-1} = \frac{1}{n!}\tilde{\epsilon}_{\mu_1\dots\mu_n}\mathrm{d}x^{\mu_1}\wedge\cdots\wedge\mathrm{d}x^{\mu_n}, \end{equation}$$ since both the wedge product and the Levi-Civita symbol are completely antisymmetric.

How do I notice that? I am struggling to see how the fact that both objects are completely antisymmetric implies the above equation. Does it mean that any completely antsymmetric tensor can be written as a contraction over the Levi-Civita symbol (I hope this is the right terminology; correct me if I'm wrong)?

A totally antisymmetric symbol in $$n$$ dimension with $$n$$ indices has only one independent component. Eg. if $$\rho_{a_1...a_n}$$ is a totally antisymmetric symbol, then $$\rho_{a_1...a_n}=\rho_{12...n}\tilde{\epsilon}_{a_1...a_n},$$ where $$\tilde\epsilon$$ is the Levi-Civita symbol. It is because in $$n$$ dimensions an antisymmetric symbol of $$k$$ indices has $$\left(\begin{matrix}n \\ k\end{matrix}\right)$$ components, and $$\left(\begin{matrix}n \\ n\end{matrix}\right)=1$$.
Which means that $$\mathrm dx^{a_1}\wedge...\wedge\mathrm dx^{a_n}=\mathrm dx^{1}\wedge...\wedge\mathrm dx^{n}\tilde{\epsilon}^{a_1...a_n}.$$ Now, we know that $$\tilde{\epsilon}_{a_1...a_n}\tilde{\epsilon}^{a_1...a_n}=n!$$, thus $$\mathrm dx^{a_1}\wedge...\wedge\mathrm dx^{a_n}\tilde{\epsilon}_{a_1...a_n}=\mathrm dx^{1}\wedge...\wedge\mathrm dx^{n}\tilde{\epsilon}^{a_1...a_n}\tilde{\epsilon}_{a_1...a_n}=n!\mathrm dx^{1}\wedge...\wedge\mathrm dx^{n},$$ which, after division by $$n!$$ produces the required formula.
This question was already answered some time ago, however I am itching to clearify something since it sometimes leads to confusion. Essentially, the Levi-Civita tensor is the volume form. They are the very same object. This becomes clear as soon as you go to a coordinate free describtion. From this point, the equation $$\epsilon=\text{d}x^0\wedge…\text{d}x^{n-1} = \frac{1}{n!} \epsilon_{\mu_1...\mu_n} \text{d}x^{\mu_1}…\text{d}x^{\mu_n}$$ is nothing but the coordinate representation of $$\epsilon$$.
This is in fact true for any fully antisymmetric tensor. To see this Consider the tensor contraction $$T^{ijk}\tilde\epsilon_{ijk}$$ with $$T$$ a fully antisymmetric tensor. For 3 dimensions we can fully expand the sum: \begin{align}\tilde\epsilon_{ijk}T^{ijk}=&\tilde\epsilon_{123}T^{123}+\tilde\epsilon_{231}T^{231}+\tilde\epsilon_{312}T^{312}+\\ &\tilde\epsilon_{132}T^{132}+\tilde\epsilon_{321}T^{321}+\tilde\epsilon_{213}T^{213}\end{align} Notice that the top row consists of even permutations and the bottom row of odd permutations. Since $$T$$ is fully antisymmetric you can equate each term to $$T_{123}$$ by permuting the indices at the possible cost of a minus sign. The top row doesn't get a minus sign since its permutations are even. The bottom row does get minus signs, but you can permute the Levi-Civita the same way to introduce another minus sign. Since $$\tilde\epsilon_{123}=1$$ you get 6 times the same term: $$\tilde\epsilon_{ijk}T^{ijk}=6\cdot T^{123}$$ Where 6 is ofcourse the number of permutations for $$n=3$$. This only works if the number of indices is equal to the dimension of the indices.